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AP Physics Electric Field Homework Problem

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  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    A has a charge of -2uC, and B has a charge of +3uC. Initially, there is no charge at point P. What is the electric field at point P due to charges A and B? Diagram given below, ignore periods; they were used solely for spacing purposes.
    ..........(0,7m)
    .............P
    .............|
    .............|
    .............|
    .............|
    A_______|________B
    (-2m,0) | (2m,0)


    2. Relevant equations

    E = kQ/r2
    C = 1x10-6 uC

    3. The attempt at a solution

    I have found the electric fields for charges A and B, but I have no clue what to do after that..

    EA= 338 N/C
    EB= 506 N/C

    Thanks in Advance for the help.
     
  2. jcsd
  3. Feb 12, 2012 #2

    cepheid

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    Welcome to PF,

    E_A and E_B are vectors. (Electric field is a vector quantity). What you've computed so far is just the magnitudes of those vectors. (I haven't checked your math to see if you've done so correctly). The directions of those vectors are:

    - in a straight line from A to P
    - in a straight line from B to P

    respectively.

    So, to get the total electric field at P, you need to find the vector sum of these two. It shouldn't be too hard, since you have all the geometry you need.
     
  4. Feb 12, 2012 #3
    Oh okay, I didn't realize that it was a vector.


    Thank you very much for the help!
     
    Last edited: Feb 12, 2012
  5. Feb 12, 2012 #4

    cepheid

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    Yes, electric field is force per unit charge. In other words, it maps out how much force a +1 C charge would experience at various points in space, as a result of other charges. So, in a sense, the electric field gives you a way of characterizing the influence of these other charges on their surroundings.

    Since force is a vector quantity, electric field, which has dimensions of [force]/[charge] is also a vector quantity. (If you divide a vector by a scalar, you still have a vector as the result).
     
  6. Feb 12, 2012 #5
    Alright that makes sense. I realized after I posted the previous question that it didn't make a difference. Thanks again!
     
  7. Feb 12, 2012 #6

    cepheid

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    I made a mistake. The direction of E_A is in a straight line from P to A (not vice versa), since charge A is negative, and the E-field is defined as the force a positive unit charge would experience. This force is therefore attractive (towards A) in this case.
     
  8. Feb 12, 2012 #7
    Alright, that is what I was thinking. It shouldn't change the magnitude though, just the direction of the electric field, correct?
     
  9. Feb 13, 2012 #8

    cepheid

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    Yeah that's correct. The magnitude only depends on the separation, r, between the charge producing the field, and the point P in space where you're evaluating that field.
     
  10. Feb 13, 2012 #9
    Alright. Thank you so much for all the help!
     
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